Is this a theorem you have made up? If so, please prove it, or give a source for the theorem and its proof.
The only theorem I have seen is f is uniformly continuous on [a,b] if f is continuous on [a,b].
We are all aware of the epsilon/delata requirement.
The OP can be proved for a finite closed interval.
None of the proofs so far are correct. Do you have a proof?
Buck, Rosenlicht, Rudin, Spivak and Taylor give the theorem for a compact space (finite, closed interval). Referring to an obscure book which most people never heard of or don't have is quite convenient.
Personally, I don't believe it, things get tricky when limits become infinite. I would be more inclined to believe it if someone could prove OP is uniformly continuous by basic definition. Perhaps you could?
However, if the theorem is indeed true, then you have answered OP.
Let $f$ be a function that is continuous on $[a,\infty)$ and $\lim_{x \to \infty} f(x) = k$.
Given $\varepsilon>0$, choose $b \in [a,\infty)$ such that for all $x \ge b$, $\left| f(x) - k \right| < \dfrac{\varepsilon}{3}$ (we know $b$ exists since the limit as $x \to \infty$ of $f(x)$ exists).
Since $f$ is continuous on the compact set $[a,b] \subseteq [a,\infty)$, by the Heine-Cantor theorem, $f$ is uniformly continuous on $[a,b]$. Choose $\delta>0$ such that given any $x,y \in [a,b]$ with $|y-x|< \delta$, $|f(y) - f(x)| < \dfrac{\varepsilon}{3}$.
Let $x,y \in [a,\infty)$ with $|y-x| < \delta$. We have the following possibilities:
1. $x\le b$ and $y\le b$
2. $x\le b$ and $y> b$ or $x>b$ and $y\le b$
3. $x>b$ and $y>b$
Case 1:
Since we already know that $f$ is uniformly continuous on $[a,b]$, we know $|f(y) - f(x)| < \dfrac{\varepsilon}{3} < \varepsilon$
Case 2:
Showing the case when $y>x$ (the case when $x>y$ is similar).
By the triangle inequality, we know
$|f(y) - f(x)| \le |f(y) - k| + |k - f(b)| + |f(b) - f(x)|$
We chose $b$ such that for all $x\ge b$, $|f(x) - k| < \dfrac{\varepsilon}{3}$. Also, we know that since $x \le b < y$ (by assumption), then $|b-x| < |y-x| < \delta$, so we have
$|f(y) - f(x)| \le |f(y) - k| + |k - f(b)| + |f(b) - f(x)| < \dfrac{\varepsilon}{3} + \dfrac{\varepsilon}{3} + \dfrac{\varepsilon}{3} = \varepsilon$
Case 3:
$|f(y) - f(x)| \le |f(y) - k| + |k - f(x)| < \dfrac{\varepsilon}{3} + \dfrac{\varepsilon}{3} < \varepsilon$
Q.E.D.
OK, given ε you can pick b. After that, I will check that the same δ applies to all cases. Intuitiveley, you shouldn't have to break it down by cases depending on where x and y are.
Thanks for giving an intelligible proof which can be followed. It's interesting. It appears that it also applies to a finite interval which makes me wonder.
I note derivative is not continuous on the interval but that is only sufficient, but not necessary (if your continuity proof is correct).
Why doesn't someone nail it by answering OP using basic definition of uniform convergence?
Edit: Whoops. Given ε, you can pick b, which depends on x. Maybe that's why the method isn't used in a standard proof for[a,b]
O.K. I will do that for you, even though I doubt you will follow it.
Start every continuity proof with: SUPPOSE $\varepsilon > 0$.
Note that $\displaystyle{\lim _{x \to \infty }}\left( {f(x) = {{\sin ({x^3})} \over x}} \right) = 0$.
That means that $\left( {\exists N > a} \right)\left[ {x \geqslant N \Rightarrow \;\left| {f(x)} \right| < \dfrac{\varepsilon }{3}} \right]$
Now $f$ is uniformly continuous on $[a,N]$ so $\left( {\exists {\delta _1} > 0} \right)\left[ {\left| {x - y} \right| < {\delta _1} \Rightarrow \;\left| {f(x) - f(y)} \right| < \frac{\varepsilon }{3}} \right]$.
Also by continuity $\left( {\exists {\delta _2} > 0} \right)\left[ {\left| {N - y} \right| < {\delta _2} \Rightarrow \;\left| {f(N) - f(y)} \right| < \frac{\varepsilon }{3}} \right]$.
Let $\delta = \min \left\{ {{\delta _1},{\delta _2}} \right\}$ so suppose that $\left| {p - q} \right| < \delta $.
Three cases: $ \begin{array}{l} 1.~\{p,q\}\subset [a,N]\\2.~p<N< q\\3.~ \{p,q\}\subset [N,\infty)\end{array} $
In each case $\left| {f(p) - f(q)} \right| < \varepsilon $.
BTW: Philip Zipse is hardly obscure.
The function f(x) = x^{2} is contiuous but not uniformly continuous on (0,infinity). See example 10 (pg 3) of:
http://www.math.wisc.edu/~robbin/521dir/cont.pdf
From the same link,f(x) is uniformly contiuous on any interval (Theorem 3, pg 3) if it satisfies a Lipschitz inequality:
│f(x)-f(x_{0})│≤ M│x-x_{0} │< Mδ = ε